Question

Choose the correct answer form the alternatives given.
If $\sin \theta +{\sin }^2\theta =1,$ then find the value of ${\cos }^{12}\theta +3{\cos }^{10}\theta +3{\cos }^8\theta +{\cos }^6\theta -1$

Answer 1

$\sin$$\theta =$$1-{\sin }^2\theta$
$\sin$ $\theta$ $={\cos }^2\theta$
Put values in the equation,
${\cos }^{12}\theta +3{\cos }^{10}\theta +3{\cos }^8\theta +{\cos }^6\theta -1$
${\sin }^6\theta +3{\sin }^5\theta +3{\sin }^4\theta +{\sin }^3\theta -1$
$({\sin }^2\theta +\sin \theta {)}^3-1$
$1-1=0$ . [$\sin \theta +{\sin }^2\theta =1$]